Tom Donaly wrote:
All of them should
go back to school and study the whole elephant, so they won't
keep making the same mistakes the three blind men made.

The following all discuss lossless systems in their
writings - Slater, Chipman, Ramo & Whinnery, Johnson,
Kraus, and Balanis. Which of those people are the
three blind men?
--
73, Cecil http://www.w5dxp.com

Tom Donaly wrote:
Cecil Moore wrote:
Good Grief, Gene, I don't have time to teach you
quantum electrodynamics. Go read a book that tells
you about the nature of photons. It is also the
cornerstone of relativity.

Cecil, you couldn't teach anyone quantum electrodynamics if
they put a gun to your head. Quit pretending.

Probably true, but Gene obviously knows less about the
subject than I do and that's pretty sad.
--
73, Cecil http://www.w5dxp.com

(Richard Harrison) wrote in news:2484-476F9B75-
:

....
No. Walter is exactly right, and don`t drag any distortionless line into
the discussion. That is a device for audio circuits. For RF, you only
need a low-loss (Zo = R) transmission line.
....

Richard, a lossless line is a special case of a distortionless line (ie all
lossless lines are distortionless lines, but not vice versa).

Specifying a distortionless line is less limiting than restriction to only
lossless lines.

You are actually proposing a greater restriction than I, and it isn't
necessary.

Owen

Dave wrote:
"Keith Dysart" wrote in message
...
On Dec 23, 10:12 am, "Dave" wrote:

Are you really prepared to throw away P = VI?

yes, when the V and I are the superimposed voltage and current that you
insist are the real current and voltage on the line.
. . .

Amazing! I suppose Ohm's law is next?

Roy Lewallen, W7EL

Dave wrote:
"Keith Dysart" wrote in message
...

Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

it is extremely simple. use traveling waves then V*I works everywhere all
the time. use standing waves and it fails. period, end of story.

It does not fail. Present any example of a transmission line terminated
with a load. Choose any line and load impedance and any transmission
line length. I'll tell you the V and I at any point on the line,
calculate the instantaneous power from V(t) * I(t) at that point, and
from that the average power. Then I'll show that this average power
equals the power in the load and the power delivered by the source. As
an added bonus, I can tell you the impedance (ratio of V/I) at any point
along the line. I'll provide both equations that always work and
numerical results.

Then you can show where I've made an error and why my calculations are
invalid.

Can I be any more fair than that?

Roy Lewallen, W7EL

Roy Lewallen wrote:
Dave wrote:
"Keith Dysart" wrote:

"Dave" wrote:
Are you really prepared to throw away P = VI?

yes, when the V and I are the superimposed voltage and current that
you insist are the real current and voltage on the line.

Amazing! I suppose Ohm's law is next?

Hopefully, you realize that P = VI only works for DC.
AC and RF, with their associated phase angles, are much
more complicated than P = VI. As a matter of fact, if
you use P = VI for standing waves, you are infinitely
wrong. There is no power in standing waves no matter
what the voltage and current magnitudes.
--
73, Cecil http://www.w5dxp.com

Roy Lewallen wrote:
Dave wrote:
"Keith Dysart" wrote:
Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

it is extremely simple. use traveling waves then V*I works everywhere
all the time. use standing waves and it fails. period, end of story.

It does not fail.

Good grief, Roy. The net voltage is 100v. The net current
is 1 amp. The load is infinite. P does NOT equal 100 watts.
P equals zero. P does not equal V * I. P = V*I*cos(A) where
A = 90 degrees.

What the hell is wrong with you people?
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
This photonic limitation is something that exists only in your head.

Good Grief, Gene, I don't have time to teach you
quantum electrodynamics. Go read a book that tells
you about the nature of photons. It is also the
cornerstone of relativity.

Cecil,

I recall from some previous messages that you may have read Feynmann's
QED. Why don't you add some Feynmann diagrams into this discussion and
edumacate us all? (I can answer that one: you wouldn't know what to do
with a Feynmann diagram if it bit you on the behind.)

More to the point. If you would stop the silly business about "net"
everything you might begin to understand that standing waves have
effects that move at the speed of light just in the same manner that
traveling waves do. Do you really believe that the fields from the
standing waves don't propagate?

Perhaps you should review the difference between static and stationary.
The standing wave is stationary. It is not static.

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:
With such profound statements as, "a pure standing wave is technically
NOT an EM wave", it you might either offer some sort of reference or
start planning your trip to Stockholm.

Gene, here is a true/false quiz for you. If you have a
reference that disagrees with the obvious answers, please
quote it.

1. Is EM wave energy photonic in nature? ________

2. Do photons move at the speed of light in a medium? ______

3. Do standing waves move at the speed of light? _______

If the answers are yes, yes, and no, then standing waves
have been eliminated from the set of EM waves.

Cecil,

Too easy.

1. Yes, but who cares? If you want to go beyond ordinary classical
models, then most of your standard formalism about transmission lines
would need a bit of work. Classical works pretty well for HF.

2. Yes.

3. Yes.


Do I win the prize?

73,
Gene
W4SZ

Roger wrote:
. . .
You give a good example Keith. It would be correct for measurement at
the load and at every point 1/2 wavelength back to the source from the
load, because the standing wave has the same measurements at these
points. At the 1/4 wavelength point back from the load and every
successive 1/2 wave point back to the source, the equation would also be
correct as demonstrated in Roy's example earlier today.

Excepting for these points, we would also be measuring a reactive
component that could be described as the charging and discharging of the
capacity or inductive component of the transmission line. (Imagine that
we are measuring the mismatched load through a 1/8 wave length long
transmission line, using an Autek RX VECTOR ANALYST instrument) The
inclusion of this reactive component would invalidate the power reading
if we were assuming that the measured power was all going to the load.
. . .

Well, let's look at that problem. Make the line 1/8 wavelength long
instead of 1/4 wavelength. The ratio of V to I at the source can be
calculated directly with a single formula or by separately calculating
the forward and reverse traveling voltage and current waves and summing
them. The result, for my 50 ohm transmission line terminated with 25
ohms, is 40 + j30 ohms. V at the input is fixed by the source at 100
volts RMS. I = V / Z = 1.6 - j1.2 amps = 2.0 amps magnitude with a phase
angle of -36.87 degrees.

Now I'll translate V and I into time domain quantities. (I could have
calculated I directly in the time domain, but this was simpler.)

Using w for omega, the rotational frequency,

If V(t) = 141.4*sin(wt) [100 volts RMS at zero degrees] then
I(t) = 2.828*sin(wt-36.87 deg.) [2 amps RMS at -36.87 degrees]

Multiplying V * I we get:

V(t) * I(t) = 400 * sin(wt) * sin(wt - 36.87 deg.)

By means of a trig identity, this can be converted to:

= 200 * [cos(36.87 deg) - cos(2wt - 36.87 deg.)]

cos(36.87 deg) = 0.80, so

V(t) * I(t) = 160 - 200 * cos(2wt - 36.87 deg.)

This is a waveform I described in my previous posting. The cosine term
is a sinusoidal waveform at twice the frequency of V and I. The 160 is a
constant ("DC") term which offsets this waveform. The fact that the
waveform is offset means that the power is positive for a larger part of
each cycle than it is negative, so during each cycle, more energy is
moved in one direction than the other. In fact, the offset value of 160
is, as I also explained earlier, the average power. It should be
apparent that the average of the first term, 160, is 160 and that the
average of the second term, the cosine term, is zero.

Let's see how this all squares with the impedance I calculated earlier.

Average power is Irms^2 * R. The R at the line input is 40 ohms and the
magnitude of I is 2.0 amps RMS, so the power from the source and
entering the line is 2^2 * 40 = 160 watts. Whaddaya know. Let's try
Vrms^2 / R. In this case, R is the shunt R. The line input impedance of
40 + j30 ohms can be represented by the parallel combination of 62.50
ohms resistive and a reactance of 83.33 ohms. The 62.50 ohms is what we
use for the R in the V^2 / R calculation. Vrms^2 / R = 100^2 / 62.50 =
160 watts.

We can also calculate the power in the load from its voltage and current
and, with the assumption of a lossless line I've been using, it will
also equal 160 watts.

P = V(t) * I(t) always works. You don't need power factor or reactive
power "corrections", or to have a purely resistive impedance.

This is really awfully basic stuff. Some of the posters here would come
away with a lot more useful knowledge by spending their time reading a
basic electric circuit text rather than making uninformed arguments.

Roy Lewallen, W7EL